Question

How do you verify $ \left( {\dfrac{{\tan x}}{{1 + \sec x}}} \right) + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) = 2\csc x $ ?

Answer 1

Hint: Here, you are asked to verify an identity which includes the trigonometric ratios $ \tan x $ , $ \sec x $ , and $ \csc x $ . What you need to do here is to use all the material or precisely trigonometric properties and identities which involves the trigonometric ratios $ \sin \theta $ , $ \cos \theta $ , $ \tan \theta $ and $ \sec \theta $ . First you consider the left-hand side and try to simplify it. What you can do, is modify the given terms and try using the trigonometric identity which gives you the relation between the tangent and secant of an angle, that is $ {\sec ^2}x - {\tan ^2}x = 1 $ . In order to proceed, cross multiply and add the first and second term on the left-hand side.

Complete step by step solution:
So, we have our left-hand side equal to $ \left( {\dfrac{{\tan x}}{{1 + \sec x}}} \right) + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) $ . Now, here what we do is, we multiply and divide by $ {\sec ^2}x - 1 $ , that is, \[\dfrac{{{{\sec }^2}x - 1}}{{{{\sec }^2}x - 1}}\]. So, basically what we are doing is just multiplying the first term by $ 1 $ , it will not change the equation or say change the first term.
So, we have,
 $
  \left( {\dfrac{{\tan x}}{{1 + \sec x}}} \right) + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) \\
   = \left( {\dfrac{{\tan x}}{{1 + \sec x}}} \right)\left( {\dfrac{{\sec x - 1}}{{\sec x - 1}}} \right) + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) \\
  $
Now, let us multiply the numerator with the numerator and denominator with denominator.
 $
  \left( {\dfrac{{\tan x}}{{1 + \sec x}}} \right)\left( {\dfrac{{\sec x - 1}}{{\sec x - 1}}} \right) + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) \\
  \dfrac{{\tan x\sec x - \tan x}}{{\sec x - 1 + {{\sec }^2}x - \sec x}} + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) \\
  \dfrac{{\tan x\sec x - \tan x}}{{{{\sec }^2}x - 1}} + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) \\
  $
Now, one of the trigonometric identities we have is $ {\sec ^2}x - {\tan ^2}x = 1 $ . From here, we will get $ {\sec ^2}x - 1 = {\tan ^2}x $ . Let us substitute this value in the denominator of the first term. We will get,
 $
  \dfrac{{\tan x\sec x - \tan x}}{{{{\sec }^2}x - 1}} + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) \\
   = \dfrac{{\tan x\sec x - \tan x}}{{{{\tan }^2}x}} + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) \;
  $
First, we will simplify our second term which is $ \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) $ In order to add $ 1 $ and $ \dfrac{{\sec x}}{{\tan x}} $ , the denominator of both should be same and that is why we will multiply and divide $ 1 $ by $ \tan x $ to make their denominator equal.
$ 1 + \dfrac{{\sec x}}{{\tan x}} = 1\left( {\dfrac{{\tan x}}{{\tan x}}} \right) + \dfrac{{\sec x}}{{\tan x}} = \dfrac{{\tan x + \sec x}}{{\tan x}} $ .
Let us get back to our previous equation and substitute
$ 1 + \dfrac{{\sec x}}{{\tan x}} = \dfrac{{\tan x + \sec x}}{{\tan x}} $ .
 $
\Rightarrow \dfrac{{\tan x\sec x - \tan x}}{{{{\tan }^2}x}} + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) \\
   = \dfrac{{\tan x\sec x - \tan x}}{{{{\tan }^2}x}} + \dfrac{{\tan x + \sec x}}{{\tan x}} \\
   = \tan x\left( {\dfrac{{\sec x - \tan x}}{{{{\tan }^2}x}}} \right) + \dfrac{{\tan x + \sec x}}{{\tan x}} \\
   = \dfrac{{\sec x - \tan x}}{{\tan x}} + \dfrac{{\tan x + \sec x}}{{\tan x}} \\
   = \dfrac{{\sec x - \tan x + \tan x + \sec x}}{{\tan x}} \\
   = \dfrac{{2\sec x}}{{\tan x}} \;
  $
Now, let us convert all the ratios involved in the simplified version of the equation given to us into sine and cosine. As we have $ \cos x = \dfrac{1}{{\sec x}} $ and $ \tan x = \dfrac{{\sin x}}{{\cos x}} $
Therefore, we have
$\Rightarrow \left( {\dfrac{{\tan x}}{{1 + \sec x}}} \right) + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) = \dfrac{{2\left( {\dfrac{1}{{\cos x}}} \right)}}{{\left( {\dfrac{{\sin x}}{{\cos x}}} \right)}} = \dfrac{2}{{\sin x}} $ .
Since $ \sin x = \dfrac{1}{{\csc x}} \to \csc x = \dfrac{1}{{\sin x}} $ , we have
$\Rightarrow \left( {\dfrac{{\tan x}}{{1 + \sec x}}} \right) + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) = 2\csc x $ .
Hence proved $ \left( {\dfrac{{\tan x}}{{1 + \sec x}}} \right) + \left( {1 + \dfrac{{\sec x}}{{\tan x}}} \right) = 2\csc x $ , identity verified.

Note: It is really important to know the trick we used here in order to simplify it, otherwise it was really hard to solve this question. You might consider the idea of multiplying and dividing it by $ {\sec ^2}x - 1 $ came from the need of getting rid from that $ 1 $ present in denominator of $ \dfrac{{\tan x}}{{1 + \sec x}} $ . The only way to get rid of that was to convert that whole term into a function of $ \tan x $ . Also, you need to memorize all the trigonometric properties such as $ {\sec ^2}x - {\tan ^2}x = 1 $ , $ \cos x = \dfrac{1}{{\sec x}} $ and $ \tan x = \dfrac{{\sin x}}{{\cos x}} $ so that it will be easy for you to solve more questions like these.